IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Others => GCSE => Topic started by: astarmathsandphysics on October 06, 2008, 11:06:45 am

Title: Maths & Physics Help
Post by: astarmathsandphysics on October 06, 2008, 11:06:45 am
Post your GCSE questions here to get the answers back asap from someone with 3 maths/physics degrees.
Title: Re: Maths & Physics Help
Post by: nid404 on April 22, 2009, 07:14:10 am
I tried starting a new topic but it's not working for me. Anyways plz help me
The second part of the question needs help.
Q.The air in a bicycle tyre has a volume of 1000cm3 and a pressure of 2.5 atmospheres. If the air is released ( at the same temperature) so its pressure is 1 atmosphere, what is the volume of the air?

This part needs help

A bicycle has a volume of 100cm3 and is now used to pump up the tyre. After 1 stroke of the pump, what is the presure in the tyre?

Thanks
Title: Re: Maths & Physics Help
Post by: spookyjaransh on November 03, 2009, 09:54:52 am
the constant here is = (1000/2.5) =400.
there fore after 1 stroke the volume increases by 100 cm3, thus new volume = 1100.
but because the constant does not change, therefore, 1100/x = 400
=> x = 1100/400 = 2.75 atm.
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on November 03, 2009, 10:26:46 am
How did I miss this question for so long
Title: Re: Maths & Physics Help
Post by: ~~~~shreyapril~~~~ on November 03, 2009, 10:39:58 am
hm..!!
i guess you did not notice it!!
u are too busy person rite??
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on November 03, 2009, 10:58:16 am
v busy
Title: Re: Maths & Physics Help
Post by: @d!_†oX!© on November 03, 2009, 12:04:23 pm
the constant here is = (1000/2.5) =400.
there fore after 1 stroke the volume increases by 100 cm3, thus new volume = 1100.
but because the constant does not change, therefore, 1100/x = 400
=> x = 1100/400 = 2.75 atm.
hey ansh!!!
u've done one thing wrong here....
according to the boyle's law constant = pressure*volume
so, k=1000*2.5 = 2500
thus,
p= k/v
=>p=2500/1100
=>p=2.27 atm

am i right sir???
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on November 03, 2009, 12:16:27 pm
You were supposed to work out the volume of the air I think when the pressure is reduced from 2.5 atm to 1.
Title: Re: Maths & Physics Help
Post by: Ghost Of Highbury on November 03, 2009, 12:17:28 pm
is it 2500/2600 = 0.9?
Title: Re: Maths & Physics Help
Post by: @d!_†oX!© on November 03, 2009, 12:18:03 pm
that was the first part...i think nid said she wanted help in the second part...
Title: Re: Maths & Physics Help
Post by: @d!_†oX!© on November 03, 2009, 12:18:28 pm
is it 0.9?
u talking abt the first or second part??
Title: Re: Maths & Physics Help
Post by: Ghost Of Highbury on November 03, 2009, 12:19:42 pm
I tried starting a new topic but it's not working for me. Anyways plz help me
The second part of the question needs help.
Q.The air in a bicycle tyre has a volume of 1000cm3 and a pressure of 2.5 atmospheres. If the air is released ( at the same temperature) so its pressure is 1 atmosphere, what is the volume of the air?

This part needs help

A bicycle has a volume of 100cm3 and is now used to pump up the tyre. After 1 stroke of the pump, what is the presure in the tyre?

Thanks

its ritten the second part needs help.
Title: Re: Maths & Physics Help
Post by: spookyjaransh on November 03, 2009, 12:24:55 pm
so wats the correct answer then?? ???
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on November 03, 2009, 12:53:43 pm
pV=k so 2.5*1000=k so V=2500/p but p is now 1atm so V=2500cm£

Some of the air escaped to the atmosphere
Title: Re: Maths & Physics Help
Post by: @d!_†oX!© on November 03, 2009, 12:55:39 pm
sir cud u do the second part as well??
Title: Re: Maths & Physics Help
Post by: Ghost Of Highbury on November 03, 2009, 12:58:50 pm
the V is 2500cm3..when 100cm3 gas is added V = 2600cm3

pressure = k/2600

k=2500

pressure = 0.9

0.9<1

pressure decreases from 1 to 0.9 volume increases from 2500 to 2600

i suppose this is the way if the second part is a continuation
Title: Re: Maths & Physics Help
Post by: @d!_†oX!© on November 03, 2009, 01:01:36 pm
see if the second part is a continuation...den this is ok...if not den mine is ok....
so how do we decide???where is nid???she might be knowing the right answer!!
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on November 03, 2009, 01:17:46 pm
The second part is comletely diff
Title: Re: Maths & Physics Help
Post by: nid404 on November 03, 2009, 01:55:29 pm
Well, I think I got this question from a gcse textbook....I think aadi it's from the one I gave you. Check kar....gas laws mein....i think it's from there.....So basically even I forgot abt this question. I really don't know what the initial volume and pressure should be taken as.

I would go for what adi's done..........

aadi check that book again
Title: Re: Maths & Physics Help
Post by: ~~~~shreyapril~~~~ on November 04, 2009, 01:00:47 pm
hm..!!
will these type of q's come in IG papers too!???
Title: Re: Maths & Physics Help
Post by: Ghost Of Highbury on November 04, 2009, 01:11:44 pm
hm..!!
will these type of q's come in IG papers too!???

y not?...its not that difficult.
Title: Re: Maths & Physics Help
Post by: ~~~~shreyapril~~~~ on November 04, 2009, 01:14:28 pm
hm..!!hope so they dont come in alt. to pracs!!

By the way watyou studying for them???

ne imp topics???
Title: Re: Maths & Physics Help
Post by: nid404 on November 04, 2009, 01:15:32 pm
abbe...woh book mein dekh...question hoga usme,...answer bhi hoga
Title: Re: Maths & Physics Help
Post by: Ghost Of Highbury on November 04, 2009, 01:30:38 pm
abbe...woh book mein dekh...question hoga usme,...answer bhi hoga
cant find it..
Title: Re: Maths & Physics Help
Post by: nid404 on November 04, 2009, 01:39:23 pm
cant find it..

chor de......mujhe yaad hi nahi phir...kaha se aaya yeh question
Title: Re: Maths & Physics Help
Post by: ~~~~shreyapril~~~~ on November 04, 2009, 04:07:21 pm
which book!!

e-book??
Title: Re: Maths & Physics Help
Post by: Tumble Bug on November 14, 2009, 01:39:04 pm
 i need help with momemntum questions!!! i have my ohy theory this monday....
:(:(:( :'(
Title: Re: Maths & Physics Help
Post by: falafail on November 14, 2009, 01:51:43 pm
i need help with momemntum questions!!! i have my ohy theory this monday....
:(:(:( :'(


aren't you doing AS level? wrong forum :P
Title: Re: Maths & Physics Help
Post by: Tumble Bug on November 14, 2009, 02:06:27 pm
sawry...my bad!!
Title: Re: Maths & Physics Help
Post by: ~~~~shreyapril~~~~ on November 17, 2009, 06:43:27 pm
dont know y all a and as level people post most of there queries in here!! :P
Title: Re: Maths & Physics Help
Post by: Q80BOY on November 26, 2009, 09:51:57 am
cause its not used much (this thread) so any posts here are noticable .. thus the many random doubts .. naughty naughty members :P :P
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on November 26, 2009, 09:53:38 am
Would be easier for me if all the maths and physics questions were in the same topics. Else I might not find them
Title: Re: Maths & Physics Help
Post by: Q80BOY on November 26, 2009, 10:00:54 am
heyy sir .. long time no see ;D

so like a similar topic (like this one) in every board?
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on November 26, 2009, 10:29:17 am
Yes |I did create one but people  make their own topics. It matters not. They are easier to find on my mobile phone so I post to say I will answer when I get home then look for my last posts when I get home.
Title: Re: Maths & Physics Help
Post by: Tumble Bug on December 20, 2009, 12:47:04 pm
please i need someone to urgently solve this question!!...integration:

the limits are 1 and 0...it is 

x.f(x)

and f(x)= 2(x+2)/5
Title: Re: Maths & Physics Help
Post by: nid404 on December 20, 2009, 02:26:09 pm
fx--> 2x+4/5
i dunno how to use the integration sign

f integration (x)--->2x/5 dx + 4/5dx
                            x2/5 +4/5x
                         
sorry i made an error in the previous integration

but u still don't get the ans......i will try agn                         
  
Title: Re: Maths & Physics Help
Post by: Tumble Bug on December 20, 2009, 02:31:14 pm
fx--> 2x+4/5
i dunno how to use the integration sign

f integration (x)--->2x2/5X2 + 4/5x
                           ---> x2+4x/ 5
substitute upper and lower limits
& then subtract lower limit from upper limit

2(1)2+ 4(1)/ 5=6/5
2(0)2+ 4(0)/ 5= 0

6/5-0=6/5

I think this is how it is to be done
                         

the answer is 8/15

:( :S
Title: Re: Maths & Physics Help
Post by: Tumble Bug on December 20, 2009, 02:33:49 pm
thank you waisay!!  :)
Title: Re: Maths & Physics Help
Post by: nid404 on December 20, 2009, 02:46:53 pm
edited my post...but still not that ans
Title: Re: Maths & Physics Help
Post by: Tumble Bug on December 20, 2009, 04:12:57 pm
edited my post...but still not that ans

nevermind lets just hope nuthin of the sort cumes in my test! :-X
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on December 20, 2009, 05:16:00 pm

xf(x)= 2x(x+2)/5=2x^2/5+4x/5 integrate to get 2x^3/15+2x^2/5
then put limits in to get (2/15+2/5).0=8/15
Title: Re: Maths & Physics Help
Post by: nid404 on December 20, 2009, 06:00:08 pm
but it's 2(x+2)/5 nd not 2x(x+2)/5
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on December 20, 2009, 07:38:49 pm
It says x.f(x)
Title: Re: Maths & Physics Help
Post by: nid404 on December 21, 2009, 03:56:37 am
It says x.f(x)
sorry my bad
Title: Re: Maths & Physics Help
Post by: Tumble Bug on December 21, 2009, 11:13:39 am
xf(x)= 2x(x+2)/5=2x^2/5+4x/5 integrate to get 2x^3/15+2x^2/5
then put limits in to get (2/15+2/5).0=8/15

omg thank you so much dude!!!
:)  :D
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on January 14, 2010, 08:54:55 am
Spam?
Title: Re: Maths & Physics Help
Post by: saifalan on May 15, 2010, 07:29:27 pm

Did similar triangles and vector crap like this a long tym ago, so culdnt solve it for my brother. So Paulie please help. I am writing down the question and then attaching the picture

In the figure, the lines ST, PQ and OR are paralle. OPS, OQT and RQS are straight line and   OR= 3a   OP= b and PS=2b and OQ= 2a+b . The underlined letters are all vectors.

Express in terms of a and b

(a)  (i) PQ          (ii)  QR        (iii)  QT      (iv)  ST           I did find (a) (i) and (a) (iii) but not the rest


(b)   (i)      the area of triangle OPQ / area of triangle  SPQ

       (ii)      the area of triangle OPQ / area of triangle OST

       (iii)     the area of triangle OPQ / area of triangle ORQ


And the answers are:

(a)      (i) 2a        (ii) a-b           (iii) 4a+2b          (iv)  6a

(b)      (i) 1/2       (ii)  1/9          (iii2/3)


Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on May 16, 2010, 09:44:02 am
here
Title: Re: Maths & Physics Help
Post by: saifalan on May 16, 2010, 10:46:22 am
hey wht did u mean by the lambda sign ??  ???
Title: Re: Maths & Physics Help
Post by: astarmathsandphysics on May 16, 2010, 11:07:45 am
lambda is a parameter
lambda and mu label the points on a line so by finding them you can find the point